Fifteen identical balls have to be put in five different boxes. Each box can contain any number of balls. The total number of ways of putting the balls into the boxes so that each box contains at least two balls is equal to

Let the balls put in the box are ${\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,{\mathrm{x}}_4\text{ and }{\mathrm{x}}_5$. We have ${\mathrm{x}}_1+{\mathrm{x}}_2+{\mathrm{x}}_3+{\mathrm{x}}_4+{\mathrm{x}}_5=15,{\mathrm{x}}_{\mathrm{i}}\ge 2$

$\text{ or }\left({\mathrm{x}}_1-2\right)+\left({\mathrm{x}}_2-2\right)+\left({\mathrm{x}}_3-2\right)+\left({\mathrm{x}}_4-2\right)+\left({\mathrm{x}}_5-2\right)=5$

$\Rightarrow {\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3+{\mathrm{y}}_4+{\mathrm{y}}_5=5,{\mathrm{y}}_{\mathrm{i}}={\mathrm{x}}_{\mathrm{i}}-2\ge 0$

The total number of ways is equal to the number of non-negative integral solutions of the last equation, which is equal to ${ }^{5+5-1}{\mathrm{C}}_5{=}^9{\mathrm{C}}_5$.